Question: Divide the following complex numbers. $\dfrac{9-i}{-4+5i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${-4-5i}$. $ \dfrac{9-i}{-4+5i} = \dfrac{9-i}{-4+5i} \cdot \dfrac{{-4-5i}}{{-4-5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(9-i) \cdot (-4-5i)} {(-4)^2 - (5i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(9-i) \cdot (-4-5i)} {(-4)^2 - (5i)^2} $ $ = \dfrac{(9-i) \cdot (-4-5i)} {16 + 25} $ $ = \dfrac{(9-i) \cdot (-4-5i)} {41} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({9-i}) \cdot ({-4-5i})} {41} $ $ = \dfrac{{9} \cdot {(-4)} + {-1} \cdot {(-4) i} + {9} \cdot {-5 i} + {-1} \cdot {-5 i^2}} {41} $ $ = \dfrac{-36 + 4i - 45i + 5 i^2} {41} $ Finally, simplify the fraction. $ \dfrac{-36 + 4i - 45i - 5} {41} = \dfrac{-41 - 41i} {41} = -1-i $